Question

What do I do now? calculus problem @mathmale

Answer 1

@mathmale @hero @Zarkon

Answer 2

\[e^{-x}\] is a decreasing function...just look at the endpoints

Answer 3

I have to do it without graphing.

Answer 4

@tkhunny @Hero

Answer 5

you don't need to graph it

Answer 6

\[f(x)=e^{-x}\]

\[f'(x)=-e^{-x}<0\] and therefore it is decreasing

Answer 7

since it is decreasing the max will be at \(x=-1\) and the min will be at \(x=1\)

Answer 8

Okay, thank you

Answer 9

What about the critical point part of the question?

Answer 10

@xapproachesinfinity

Answer 11

e^-x is strictly decreasing so there is no critical points

Answer 12

remember crtical point occur when the g' changes sign
but here -e^-x <0 all the time no sign change

Answer 13

i mean extrema no critical points in general

Answer 14

I thought a criticall point was where g'(x) = 0

Answer 15

no g' is never zero, and yes critical points are found with g'=0
or g' does not exist

Answer 16

here g' is defined for all x
and g' does not equal zero
hence no critical points

Answer 17

oh one second i didn't see -1<=x<=1

Answer 18

there is max and min in this case

Answer 19

g(-1) and g(1) are max and min respectively

Answer 20

but still we say critical point don't exist

Answer 21

Okay, so what would I say to answer the second part of the question?

Answer 22

if we are just talking about extrema
absolute max and min exist

Answer 23

the question asked you for extrema first
critical points don't exist for second part

Answer 24

stationary point are critical points where g'=0
i just learned that know lol :)

Answer 25

i mean i never heard stationary before

Answer 26

Ok, thank you

Answer 27

welcome :)